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We are given a set of four points: $A=(5,-3) $ $B=(2,1)$ $C=(-5,2)$ $D=(-9,-1)$ Find the equation of the circle circumscribed on the trapezoid.

My idea to solve this problem is to simply substitute these points to the equation of a circle and solve the system of equations. $$(5-x)^2+(-3-y)^2 =R^2$$ $$(2-x)^2 + (1-y)^2 =R^2$$ $$(-5-x)^2+(2-y)^2 = R^2$$ $$(-9-x)^2+(-1-y)^2=R^2 $$ However, this system only has three variables, thus one point is unnecessary. Apart from that, this will be a long and tedious calculation. Could you suggest me a better way to tackle this?

ILoveChess
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  • If the four points are concyclic, you can find the centre by finding the intersection point of any two perpendicular bisectors of the sides. – David Quinn Apr 24 '17 at 21:32
  • find the circle that circumscribes the triangle $ABC$ , if D lies on the circle, you have your answer, if not, no such circle exists. – WW1 Apr 24 '17 at 21:53
  • @WW1 How to prove that a circle which circumscribes the triangle is the same which circumscribes the trapezoid? – ILoveChess Apr 24 '17 at 21:56
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    The circle passing through the points $A, B, C$ is unique.

    If your problem has a solution it must be that circle. Plug in $(-9,-1)$ into the equation of the circle, if it satisfies the equation then you are done, if not, your problem has no solution.

     – WW1 Apr 24 '17 at 22:13
  • In the computer programs I've seen for the circumcenter, the circle usually hits only three points. – Cye Waldman Apr 24 '17 at 22:59
  • Since a circumcircle is constructed from two perpendicular bisectors each passing through the triangle's midpoints, there is only one unique point in which they intersect, and hence only one unique centre for the circle. Also, there can only be one radius connecting the three points (they lie on the circle and have the same radius). – Toby Mak Apr 24 '17 at 23:20

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